Question

Find the eigenvalues and eigenvectors of the following matrices.\(\left(\begin{array}{rrr}-3 & 2 & 2 \\ 2 & 1 & 3 \\ 2 & 3 & 1\end{array}\right)\)

Short answer

Solve the characteristic polynomial for eigenvalues. Then, for each eigenvalue, solve \((A - \lambda_i I)x = 0\) for eigenvectors.

Step-by-step solution

- Set Up Characteristic Equation

To find the eigenvalues of matrix \(A\), first calculate the characteristic polynomial. The characteristic polynomial is given by the determinant equation \( \det(A - \lambda I) = 0 \), where \( \lambda \) represents the eigenvalues and \(I\) is the identity matrix.

- Substitute Matrix and Identity Matrix

Substitute the given matrix \(A\) and the identity matrix \(I\) into the equation: \[A - \lambda I = \begin{pmatrix}-3 & 2 & 2 \ 2 & 1 & 3 \ 2 & 3 & 1\end{pmatrix} - \lambda \begin{pmatrix}1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1\end{pmatrix} = \begin{pmatrix}-3 - \lambda & 2 & 2 \ 2 & 1 - \lambda & 3 \ 2 & 3 & 1 - \lambda\end{pmatrix}\].

- Compute the Determinant

Evaluate the determinant of the resulting matrix to get the characteristic polynomial. Compute \(\det(\begin{pmatrix}-3 - \lambda & 2 & 2 \ 2 & 1 - \lambda & 3 \ 2 & 3 & 1 - \lambda\end{pmatrix})\).

- Solve for Eigenvalues

Solve the characteristic polynomial equation for \( \lambda \). This will give you the eigenvalues of the matrix. The determinant calculation yields the polynomial equation:\[(-(3+\lambda))((1-\lambda)((1-\lambda)) - 9) - 2((4-3\lambda) - 8) + 2((3-9) - 2(-\lambda+3)) = 0\].Simplifying, we get the polynomial \[ -\lambda^3 + \lambda^2 + 10\lambda - 22 = 0\].This equation can be solved to find the eigenvalues.

- Find Eigenvectors

For each eigenvalue \( \lambda_i \), solve the equation \((A - \lambda_i I)x = 0\) to find the eigenvector associated with \( \lambda_i \). For example, if \( \lambda_1 = 2 \), substitute \( \lambda_1 \) into \(A - \lambda_1 I\) and solve for vector \(x\). Repeat this process for each eigenvalue.

Key concepts

Characteristic Polynomial

In linear algebra, a **characteristic polynomial** plays a crucial role in finding eigenvalues of a matrix. To derive the characteristic polynomial, you need to set up the characteristic equation given by \(\det(A - \lambda I) = 0 \), where \(A\) is your given matrix and \(\lambda \) represents the eigenvalues you seek. Here, \(I\) is an identity matrix—essentially a matrix in which all diagonal elements are 1 and all off-diagonal elements are 0.
To form the characteristic polynomial, subtract \(\lambda I\) from the given matrix \(A\), and calculate the determinant of the result. This determinant becomes the characteristic polynomial you solve for the eigenvalues.
For our given example, the matrix \(A\) is \(\begin{pmatrix}-3 & 2 & 2 \ 2 & 1 & 3 \ 2 & 3 & 1\textbackslashend{pmatrix}\) and the identity matrix \(I\) is \(\begin{pmatrix}1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1\textbackslashend{pmatrix}\). Calculate \(A - \lambda I\), and then find the determinant.

Determinant

The **determinant** of a matrix is a scalar value that can tell us if a matrix is invertible and is used in numerous calculations in linear algebra. To find the eigenvalues, you need to calculate the determinant of the matrix \(A - \lambda I\).
For our matrix \(A\), after substituting and simplifying, we get: \(\begin{pmatrix}-3 - \lambda & 2 & 2 \ 2 & 1 - \lambda & 3 \ 2 & 3 & 1 - \lambda \textbackslashend{pmatrix}\).
The next step is to compute the determinant of this matrix, which involves expanding along any row or column. The simplified form of \(\textbackslash det (A - \lambda I)\) gives us a polynomial in \(\lambda \), known as the characteristic polynomial. In our case, it simplifies to \(-\lambda^3 + \lambda^2 + 10\lambda - 22 = 0 \).

Matrix Algebra

In **matrix algebra**, operations like addition, subtraction, and finding determinants are essential. It's the foundation for solving systems of linear equations and performing transformations.
When you work with eigenvalues and eigenvectors, you often need to manipulate matrices. For example, given the matrix \(A\), you create another matrix by subtracting \(\lambda I\) and then calculate the determinant.
Once you have the characteristic polynomial, solving it gives you the eigenvalues. Each eigenvalue corresponds to a specific transformation described by the matrix \(A\). For our matrix, the polynomial \(-\lambda^3 + \lambda^2 + 10\lambda - 22 = 0\) reveals the eigenvalues upon solving.

Linear Equations

To find the **eigenvectors**, you need to solve systems of **linear equations**. Given each eigenvalue \( \lambda_i \), you form a new matrix \(A - \lambda_i I \).
By setting up the equation \( (A - \lambda_i I)x = 0 \), where \(x \) is the eigenvector, you solve for \( x \). This is essentially solving a homogeneous system of linear equations.
For each eigenvalue from the polynomial \(-\lambda^3 + \lambda^2 + 10\lambda - 22 = 0 \), substitute back into \(A - \lambda_i I \) and solve. This gives you the eigenvectors. For example, if \( \lambda_1 = 2 \), plug it into \(A - 2I \) and solve the resulting system to find the corresponding eigenvector. Repeat this for all eigenvalues.